無聊地想 linear algebra & independent study 的 presentation

大概在上一個星期的 numerical analysis (MATH3312/231) 課 professor 毛毛 提到了一個 fact：

Theorem. Let $A$ be real $n\times n$ matrix, then $\lim_{n\to\infty}A^n=0\iff\rho(A)<1$.

其中當 $A$ 為 $n\times n$ real matrix，我們定義 $$\sigma(A):=\{\lambda\in\mathbb{C}:A-\lambda I\text{ is not inveritble}\}$$ 為 $A$ 的 spectrum (也就是說 $\sigma(A)$ 是 $A$ 所有的 eigenvalue)。且定義 spectral radius 為 $\rho(A)=\max\{|\lambda|:\lambda \in \sigma(A)\}$。

上面 $\lim_{n\to\infty}A^n=0$ 可理解為 $\lim_{n\to\infty}A^n \mathbf x = 0,\forall \mathbf x\in \mathbb{R}^n$。Google 一下 spectral radius 這個字就有以上那個結果的證明，且用到了 Jordan decomposition，當然我有一個更簡單的解 (TA 也確定是正確的證明)，不須用到這工具。但看來就是沒有人有興趣~，還有人說這是 trivial 的 (那是一位 applied math 學生)，我無言了。

我們現在證明上述定理。可知 $(\Rightarrow)$ 是顯然的，現證 $(\Leftarrow)$：

Proof. It is a well-known result in linear algebra that every matrix can be made into an upper triangular matrix  under a change of basis. That is, by reviewing $A$ as a linear operator from $\C^n$ to $\C^n$ rather than a real matrix, there must be a basis in $\C^n$ so that the matrix of $A$ w.r.t. this basis becomes upper triangular. A simple proof can be found in ALGEBRA written by Michael Artin.

We upper triangularize $A$  by some $P\in GL_n(\C)$, i.e.,
$$U:=PAP^{-1}= \begin{bmatrix} \lambda_1& b_{12} &\cdots   &   b_{1n} \\ 0 &\lambda_2& \cdots  &  b_{2n}     \\ 0 & 0 & \ddots  &   \vdots    \\ 0 &0  &    \cdots   &\lambda_n \end{bmatrix},$$ then $U\mathbf e_1=\lambda_1 \mathbf e_1$, and $U^j \mathbf e_1=\lambda_1^j\mathbf e_1$ and hence  $|\lambda_1|<1\implies U^k \mathbf e_1\to 0$.

We complete the proof by induction, assume there is $k\in \N$ so that
$$\lim_{j\to\infty} U^j(\mathbf e_1),\dots,\lim_{j\to\infty} U^j(\mathbf e_{k-1})=0.$$ Then since $U(\mathbf e_{k})=\sum_{i=1}^{k-1} b_{ik}\mathbf e_i + \lambda_k \mathbf e_k$, we have
$$U^{j+1}(\mathbf e_{k})=\sum_{i=1}^{k-1} b_{ik}U^{j}(\mathbf e_i) + \lambda_k U^j(\mathbf e_k).$$ For a vector $v\in \C^n$ let's denote $v^\ell$ to be its $\ell^\text{th}$ coordinate, then one has
$$(U^{j+1}(\mathbf e_{k}))^\ell=\sum_{i=1}^{k-1} b_{ik}(U^{j}(\mathbf e_i))^\ell + \lambda_k (U^j(\mathbf e_k))^\ell,$$ this implies
$$\big|(U^{j+1}(\mathbf e_{k}))^\ell\big|\leq\sum_{i=1}^{k-1} |b_{ik}|\big|(U^{j}(\mathbf e_i))^\ell\big| + |\lambda_k| \big|(U^j(\mathbf e_k))^\ell\big|.$$ By induction hypothesis $\lim_{j\to\infty}U^j (\mathbf e_i)=0$ for $i=1,2,\dots,k-1$, one has for any $\epsilon>0$, there is an $N$ such that $$j>N\implies \big|(U^{j+1}(\mathbf e_{k}))^\ell\big|<\epsilon+  |\lambda_k| \big|(U^j(\mathbf e_k))^\ell\big|.$$ This actually implies $\lim_{k\to\infty} \big|(U^{j+1}(\mathbf e_{k}))^\ell \big|=0$ since $|\lambda_k|<1$. As this is true for $\ell=1,2,\dots,n$, so $\lim_{j\to\infty} U^j(\mathbf e_k)=0$.

We conclude by induction that $\lim_{j\to\infty} U^j(\mathbf e_k)=0$ for $k=1,2,\dots, n$. Since each vector in $\C^n$ is spanned by $\{\mathbf e_i\}_{i=1}^n$, we conclude $U^j\to 0$. Since $A^j=P^{-1}U^j P$, we conclude $A^j\to 0$ on $\C^n$, and of course, on $\R^n\subseteq \C^n$.$\qed$

Remark. For the ($\Leftarrow$) we may also use a famous result in Banach algebra. It is known that the formula of spectral radius of the matrix $A$ is $\limn \|A^n\|^{1/n}$. So $\limn \|A^n\|^{1/n} < 1$ implies we can choose $r<1$ such that there is $N\in\N$, $$n>N\implies \|A^n\| < r^n \implies \limn \|A^n\|=0 .$$ Finally for each $x\in \R^n$, $\|A^n x\|\leq \|A^n\| \|x\|$, so $A^n \to 0$ pointwise on $\R^n$.

---

話說我在這個 sem 跟 kin li 學 Fourier analysis。初頭也是正正常常地學 fourier series on $\mathbb{R}/\mathbb{Z}$，其後因為我在修讀的 MATH5111/511 教 finite group representation 的關係，我便開始讀 noncommutative 的 harmonic analysis 了...。最後我的 presentation project 更差點變成 existence of Haar measure on locally compact metrizable group。我其後在想，這和 fourier analysis 關係不太大吧...，便要求只證 existence of Haar measure on compact group，取而代之以 fourier series 的方法 determine $L^2(Q,m)$  的 dual，其中 $Q\subseteq\mathbb{R}$ 是 measurable 及 $m$ 是 Lebesgue measure。